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Download PROOFS Math 174 May 2017 I. Introduction. In the natural sciences
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PROOFS Math 174 May 2017 I. Introduction. In the natural sciences, we cannot prove anything to be true. We can only support hypotheses with data and make decisions based on the most strongly-supported hypotheses, but the fact remains that fringe philosophical problems still apply (can we trust our senses? Are we living in a simulation?) The great thing about mathematics is that we can prove statements, called theorems (or propositions or lemmas depending on their place in the hierarchy), to be true. II. Direct proofs. Most theorems have the form p → q, where p is some assumption we make (even one as seemingly innocent as “numbers exist”). We apply other theorems—the laws of algebra, or facts about whole numbers, or other statements we’ve proven true—and arrive at the fact that q is true. Here’s an example. Theorem. If n is an even integer, then so is n2 . This is a p → q statement where p is the statement that n is some even integer and q is the statement that n2 is an even integer. Before we dive into the proof, though, we must be comfortable with the definitions. (In fact, writing proofs is a better way to learn definitions and theorems than memorizing them!) An integer is a whole number, e.g. 21, −4, 0. An even integer is one that can be divided by 2 to yield another integer. In other words, Definition. Let n be an integer. If n is even, then there exists an integer m such that n = 2m. If n is odd, then there exists an integer m such that n = 2m + 1. (WARNING: n and m are different integers. Don’t let me catch you writing n = 2n + 1, because that equation is never true!) Definitions don’t need to be proven—we take them for granted. However, we need to be sure our audience agrees with our definitions. (Most definitions in math, especially at the undergraduate level, are standard, and everyone uses the same ones.) Now that we know what integers and even integers are, we can try to prove the theorem. In blue, I will explain each step. Proof. Let n be an even integer. By definition, there exists another integer m where n = 2m. state assumption unpack definition Now what? We are trying to prove something about n2 , so let’s square n. Be careful that you apply algebra correctly. Therefore, n2 = (2m)2 = 4m2 = . . . To prove n2 is even, we need to be able to factor out a 2. Can we? square n . . . = 2(2m2 ). Since m is a whole number, so is 2m2 . Hence, n2 is 2 times an integer. conclusion Thus if n is an even integer, then so is n2 . restate theorem III. Proofs by contradiction. Sometimes, no matter how hard you try, you cannot get from p to q with the normal methods. Maybe, worse, the theorem relies on a “negative” definition that only says what something is not. In this case, we’ll have to rely on some logical trickery. Recall from the homework that the statement [q ∧ (¬p → ¬q)] → p is a tautology. (If you haven’t done this problem yet, go do it!) Here, p is a statement we’d like to prove and q is something we know to be true. It turns out that if we assume ¬p—the opposite of what we want to prove!—and derive ¬q, we come to a contradiction. We can’t have both q and ¬q true simultaneously. What caused the problem? It was assuming ¬p, and hence p must have been true all along. Theorem. If n is an integer and n2 is even, then n is even. This theorem is the converse of the first theorem we proved. To prove it, we will assume that n is odd and derive a contradiction: namely, that n2 is odd, when it was supposed to be even. Proof. Let n be an integer such that n2 is even. Suppose n is odd. There exists an integer m where n = 2m + 1. Thus, n2 = (2m + 1)2 = 4m2 + 4m + 1. Since 4m2 + 4m is even, n2 is odd. This contradicts the assumption that n2 is even. Thus, if n is an integer and n2 is even, n must be even too. state assumption set up contradiction definition apply algebra correctly! get contradiction conclusion IV. Biconditional proofs. A theorem including the phrase “if and only if,” “is necessary and sufficient for,” “are equivalent,” etc. is a biconditional p ↔ q. These are proven by proving p → q and q → p together. Theorem. Let n be an integer. Then n is even if and only if n2 is even. We have shown that if n is even then n2 is even, and we have shown that if n2 is even then n is even. Together, they prove that n is even if and only if n2 is even. V. Irrational numbers exist. We are now prepared to prove an important theorem in mathematics: namely, that not all real numbers are rational. Theorem. The positive square root of 2 is irrational. √ Again, let’s work out the definitions.√The positive square root of 2, denoted 2, is the positive number whose square is 2: ( 2)2 = 2. A rational number is the quotient p/q of two integer p and q where q 6= 0, and if we like we may take p and q to have no factors in common so that the quotient is in lowest terms. Finally, an irrational number is one that isn’t rational. Notice that the core word in the theorem, √ “irrational,” is a negative one: “not rational.” The only course of action is to suppose that 2 is rational and come up with a contradiction. Let’s get started! Proof. There is no assumption in this proof, so we just start with the contradiction right out the gate. √ Suppose that 2 is rational. √ By definition, there exist integers p and q where q 6= 0 and 2 = p/q. Additionally, we assume p/q is in lowest terms. Again, we’ve hit a wall. The only thing we know about use that and see what happens. √ assume ¬p definition it doesn’t hurt! 2 is that its square is 2, so let’s If we square both sides, we see 2 = p2 /q 2 . We can rearrange this equation into p2 = 2q 2 , which tells us that p2 is even. Thus, p is even, too. earlier theorem Let’s stop and taken inventory of what we know: p is even, p/q is in lowest terms, and somewhere lurks a contradiction. It’s going to turn out that q is even, too; and since p and q are both even, p/q isn’t in lowest terms as assumed. This is our contradiction. Let’s see how it comes up. Because p is even, there exists an integer m with p = 2m. Substituting p = 2m into p2 = 2q 2 , we get (2m)2 = 2q 2 ⇒ 4m2 = 2q 2 ⇒ 2m2 = q 2 . Hence, q 2 and thus q are even. However, we assumed that p/q was in lowest terms, so p and q had no common factors. Therefore, p and q cannot both be even. √ By assuming 2 = p/q in lowest terms, we arrived at the conclusion that both p and q are √ even. This is contradictory. It must thus be impossible that 2 = p/q. Therefore by contradiction, no such p and q exist, and so √ 2 is irrational.